# "Spikes" on an uncalibrated oscilloscope probe, and an RC circuit question.

#### Braeden Hamson

Feb 18, 2016
237
Sorry, I don't really know how to google this. So I was using an oscilloscope in my school's lab today, our lab procedure had us clip our probes to the scopes 1 kHz calibration wave form. I know this is to get rid of what I've heard being called "batman's ears" see my picture. This picture is from a Fourier series MATLAB script not an actual scope output but it has the peaks I'm talking about. I'm wondering why this is happening. And what adjusting the screw on my probe does.

Maybe I'll take a stab at it to see if I get it right. I think I've heard this having to do with probe capacitance. Perhaps it something like the electrons are flowing through the probe and are being "pushed" along by the signal source. In the case of a square wave the signal is suddenly turned off and those electrons continue to flow for a very short time. A little bit like a water hammer. Is this accurate?

And if you feel like answering I have a second question brought about by my lab today. So I was testing an RC circuit with a square wave pulse signal. I think it was a .2 ms pulse ever 20 ms. The signals looked pretty good I had the exponential rise and fall. Then I got curious and added an offset to my signal, I added a negative offset so the signal was split -2.5v and 2.5v. And nothing happened. I mean my signal moved down on the scope screen but the wave forms didn't change. The capacitor was still discharging and charging in the same exponential way according to my scope. This makes no sense to me.

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#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,743
"batman's ears"
The correct term is ringing. A more detailed explanation you can read e.g. here.

This makes no sense to me.
I makes perfect sense.
Let's call your original signal, i.e. the square wave, y1(t) where t is time and y 1(t) is any mathematical function describing the input voltage to the filter as a function of time.
Let's call the offset y2(t). Obviously y2(t) = const = -2.5 V in your example.
The complete input signal to the filter can now be described as yin(t) = y1(t) + y2(t).
The filter has a transfer function yout(t) = F(yin(t)) where F() is the transfer function or transfer characteristic. As long as we deal with linear systems - an RC lowpass is a linear system - the math is straightforward as
F(a+b) = F(a)+F(b) or matched to above definitions yout(t) = F(y1(t)) + F(y2(t))
The term F(y1(t)) gives the rounded square wave signal. The term F(y2(t)) gives a constant offset as for an RC low pass filter the transfer characteristic for a DC signal (offset) is unity -> F(-2.5 V) = -2.5 V

Read this up e.g. here or here.

#### Braeden Hamson

Feb 18, 2016
237
Sorry for the lack of response. Been busy. Thank you for the response!

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